# Fundamentals of Rock Mechanics, Fourth Revised Edition by John Jaeger, N. G. Cook, Robert Zimmerman

By John Jaeger, N. G. Cook, Robert Zimmerman

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Extra info for Fundamentals of Rock Mechanics, Fourth Revised Edition

Example text

33) that the two principal stress directions are rotated by 90◦ from each other. The orthogonality of the principal directions continues to be the case in three dimensions; this can be proven as follows. 73) u v = ux uy uz vy  = ux vx + uy vy + uz vz = u · v. vz Also, recall that the transpose operation has the properties that (AB)T = BT AT , and (AT )T = A. Consider two principal stresses, σ1 = σ2 , and the principal directions corresponding to these two stresses, that is, τn1 = σ1 n1 and τn2 = σ2 n2 .

For example, at point P, when θ = 0, there is no rotation from the σ1 direction, and indeed Mohr’s circle indicates that (σ = σ1 , τ = 0). Similarly, consider the plane for which θ = 90◦ . This plane is rotated counterclockwise by 90◦ from the σ1 direction, and therefore represents the σ2 direction. This plane is represented on Mohr’s circle by the point that is rotated clockwise by 2θ = 180◦ , which is point Q on Fig. 6a, where we ﬁnd (σ = σ2 , τ = 0). This construction also clearly shows that the maximum shear stress has a magnitude equal to the radius of the Mohr’s circle, and occurs on planes for which 2θ = ±90◦ , which is to say θ = ±45◦ .

147). These eight planes are each parallel to a side of an octahedron whose vertices are located on the principal axes (Nadai, 1950, p. 105). 86), the normal traction acting on any octahedral plane is σoct = σ1 n12 + σ2 n22 + σ3 n32 = 1 1 (σ1 + σ2 + σ3 ) = I1 . 88), the magnitude of the shear traction acting on an octahedral plane is found to be given by √ 1 2 2 1/2 2 2 2 1/2 |τoct | = (σ1 −σ2 ) +(σ2 −σ3 ) +(σ3 −σ3 ) I +3I2 = . 149) 3 3 1 The octahedral √ shear stress has the interesting physical interpretation of being equal to (5/3) times the root-mean-square shear stress, with all planes weighted equally.